find-the-value-of-0-dx-1-x-2-1-x-4-

Question Number 33128 by prof Abdo imad last updated on 10/Apr/18

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + x^{4})} .

Commented by prof Abdo imad last updated on 13/Apr/18

l e t p u t I = \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + x^{4})} w e h a v e

2 I = \int_{- \infty}^{+ \infty} \frac{d x}{(1 + x^{2}) (1 + x^{4})} . l e t i n t r o d u c e t h e

c o m p l e x f u n c t i o n φ (z) = \frac{1}{(1 + z^{2}) (1 + z^{4})}

φ (z) = \frac{1}{(z - i) (z + i) (z^{2} - i) (z^{2} - (- i))}

= \frac{1}{(z - i) (z + i) (z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (z + \sqrt{- i})}

\sqrt{i} = e^{i \frac{π}{4}} \sqrt{- i} = e^{- i \frac{π}{4}} s o t h e p o l e s o f ? φ a r e

i, - i, e^{i \frac{π}{4}}, - e^{i \frac{π}{4}}, e^{- i \frac{π}{4}}, - e^{- i \frac{π}{4}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (R e s (φ, i) + R e s (φ, e^{i \frac{π}{4}}) + R e s (φ, - e^{- i \frac{π}{4}})

l e t p (z) = (1 + z^{2}) (1 + z^{4})

p (z) = 1 + z^{4} + z^{2} + z^{6} \Rightarrow p^{^{'}} (z) = 2 z + 4 z^{3} + 6 z^{5}

R e s (φ, i) = \frac{1}{p^{^{'}} (i)} = \frac{1}{2 i + 4 i^{3} + 6 i^{5}} = \frac{1}{2 i - 4 i + 6 i}

= \frac{1}{4 i}

R e s (φ, e^{i \frac{π}{4}}) = \frac{1}{p^{^{'}} (e^{i \frac{π}{4}})} = \frac{1}{2 e^{i \frac{π}{4}} + 4 e^{i \frac{3 π}{4}} + 6 e^{i \frac{5 π}{4}}}

b u t 2 e^{i \frac{π}{4}} + 4 e^{i \frac{3 π}{4}} + 6 e^{i \frac{5 π}{4}}

= 2 e^{i \frac{π}{4}} - 4 e^{- i \frac{π}{4}} - 6 e^{i \frac{π}{4}} = - 4 (e^{i \frac{π}{4}} + e^{- i \frac{π}{4}})

= - 4 . 2 . \frac{\sqrt{2}}{2} = - 4 \sqrt{2}

R e s (φ, - e^{- i \frac{π}{4}}) = \frac{1}{p^{^{'}} (- e^{- i \frac{π}{4}})}

= \frac{1}{- 2 e^{- i \frac{π}{4}} - 4 e^{- i \frac{3 π}{4}} - 6 e^{- i \frac{5 π}{4}}}

= \frac{- 1}{2 e^{- i \frac{π}{4}} + 4 e^{- i \frac{3 π}{4}} + 6 e^{- i \frac{5 π}{4}}} = \frac{- 1}{2 e^{- i \frac{π}{4}} - 4 e^{i \frac{π}{4}} - 6 e^{- i \frac{π}{4}}}

= \frac{- 1}{- 4 (e^{i \frac{π}{4}} + e^{- i \frac{π}{4}})} = \frac{1}{4 . 2 . \frac{\sqrt{2}}{2}} = \frac{1}{4 \sqrt{2}} \Rightarrow

2 I = 2 i π (\frac{1}{4 i} - \frac{1}{4 \sqrt{2}} + \frac{1}{4 \sqrt{2}}) = \frac{π}{2} \Rightarrow

I = \frac{π}{4} .

Commented by prof Abdo imad last updated on 13/Apr/18

a n o t h e r m e t h o d

t h e c h . x = \frac{1}{t} g i v e

I = - \int_{0}^{\infty} \frac{1}{(1 + \frac{1}{t^{2}}) (1 + \frac{1}{t^{4}})} \frac{- d t}{t^{2}}

= \int_{0}^{\infty} \frac{t^{4} d t}{(1 + t^{2}) (1 + t^{4})} = \int_{0}^{\infty} \frac{1 + t^{4} - 1}{(1 + t^{2}) (1 + t^{4})} d t

= \int_{0}^{\infty} \frac{d t}{1 + t^{2}} - \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) (1 + t^{4})} = \frac{π}{2} - I \Rightarrow

2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4}

★ I = \frac{π}{4} ★